To find the nth Derivative of $9\sqrt{x}$

Question

Can you help me to proof that the nth derivative of $9\sqrt{x}$ is $$ (-1)^{(n-1)} \cdot \frac{9(2n-2)!}{(n-1)!} \cdot (4x)^{\frac{1-2n}{2}}$$

I've tried induction but didn't go very far.

Many thanks

Answer 1

Induction is a very good way. For the induction step,

Say $P(n)$ is true for $n=k$
For $P(k+1)$, $$\frac{d}{dx^{k+1}}(9\sqrt x)$$ $$=\frac{d}{dx}\left(\frac{d}{dx^{k}}(9\sqrt x)\right)$$ $$=\frac{d}{dx} \left[(-1)^{(k-1)} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot (4x)^{\frac{1-2k}{2}}\right]$$ $$=(-1)^{(k-1)} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot \frac{d}{dx} \left[(4x)^{\frac{1-2k}{2}}\right]$$ $$=(-1)^{(k-1)} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot \frac{1-2k}{2} (4x)^{\frac{-1-2k}{2}}\cdot 4$$ $$=(-1)^{[(k+1)-1]} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot (2k-1) (4x)^{\frac{-1-2k}{2}}\cdot 2$$ $$=(-1)^{[(k+1)-1]} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot (2k-1) \cdot \frac{2k}{k} \cdot (4x)^{\frac{-1-2k}{2}}$$ $$=(-1)^{[(k+1)-1]} \cdot \frac{9[2(k+1)-2]!}{[(k+1)-1]!} \cdot (4x)^{\frac{1-2(k+1)}{2}}$$

$P(k+1)$ is true.

Answer 2

Firstly, the 9 is unimportant. Induction is the correct way to proceed, but perhaps we can disguise it in a computation:

$$\begin{align} \frac{d^n}{dx^n} \left( x^{\frac{1}{2}} \right) &= \frac{d^{n-1}}{dx^{n-1}} \left( \frac{1}{2} x^{\frac{1}{2} -1} \right) \\&= \frac{d^{n-2}}{dx^{n-2}}\left( \frac{1}{2} \left(\frac{1}{2} - 1 \right) x^{\frac{1}{2} -2} \right) \\&= \dots =\frac{d^{n-k}}{dx^{n-k}} \left( \frac{1}{2} \left( \frac{1}{2} -1\right) \left( \frac{1}{2} -2 \right)\dots \left(\frac{1}{2} -(k-1) \right) x^{\frac{1}{2} - k}\right) \\&= \frac{1}{2} \left( \frac{1}{2} - 1 \right) \left( \frac{1}{2} -2 \right)\dots \left(\frac{1}{2} - (n-1) \right) x^{\frac{1}{2} -n }. \end{align} $$ Can you simplify this?

Answer 3

Writing the first four derivatives of $f(x)=x^{1/2}$ reveals

$f^{(1)}(x)=\frac12 x^{-1/2}$

$f^{(2)}(x)=\frac12 \left(-\frac12\right)x^{-3/2}$

$f^{(3)}(x)=\frac12 \left(-\frac12\right)\left(-\frac32\right)x^{-5/2}$

$f^{(4)}(x)=\frac12 \left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)x^{-7/2}$

We propose the general term is given by

$$f^{(n)}(x)=(-1)^{n-1}\left(\frac12\right)^n (2n-3)!!x^{-(2n-1)/2}$$

Then, the $n+1$ derivative

$$\begin{align} f^{(n+1)}(x)&=(-1)^{n-1}\left(\frac12\right)^{n}(-1)\left(\frac12\right) (2n-1)(2n-3)!!x^{-(2n-3)/2}\\\\ &=(-1)^n\left(\frac12\right)^{n+1}(2n-1)!!x^{-(2n-3)/2} \end{align}$$

which provides proof by induction!